More answers

anonymous

5 years ago

Thank you!

UnkleRhaukus

5 years ago

prove by contradiction ,
assuming the square root of two is rational ,
it can be written as a ratio of natural numbers that have no common factors p,q
say
\[√2=p/q\]
\[2=p^2/q^2\]\[2q^2=p^2\]
this implies \(p^2\) is even
if \(p^2\) is even then \(p\) is even
if \(p \) is even
it can be written as
two times another natural number \(r\)
\[p=2r\]
\[2q^2=p^2=(2r)^2=4r^2\]\[q^2=2r^2\]
which similarly implies that \(q\) is even
With \(p,q\) both even the natural numbers necessarily
have a common factor of two , which contradicts the statement of rationality,
\[\sqrt2\not\in\text{rational numbers}\]
\[\square\]

anonymous

5 years ago

So in order to be rational the none of them should be even or at least one of them should be even and the other not even in order not to have a common factor of 2?
Can you please explain this situation to me if you may?

UnkleRhaukus

5 years ago

if the numer is rational
there must be a way to express it
as a ratio of (relatively prime) natural numbers, (ie no common factors)